# Prove that symplectic Lie algebras, sp(n)\mathfrak{sp}(n), are simple

The symplectic lie algebra defined by $sp\left(n\right)=\left\{ X\in gl_{2n}\,|\, X^{t}J+JX=0\right\}$ when $J=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}$. So $X\in sp\left(n\right)$ is of the sort $X=\begin{pmatrix}A & B\\ C & -A^{t}\end{pmatrix}$ when $B=B^{t},\, C=C^{t}$.

This far I was able to get, but how can I prove that it is simple?

Thanks!

Let ${\mathfrak g}=sp(n)$.

Short answer, from an advanced viewpoint : $\mathfrak g$ is semisimple (because $K(x,y)=tr(xy)$ defines a nondegenerate invariant bilinear form on $\mathfrak g$), so it will be simple iff its Dynkin diagram is connected. But we know that the diagram is $C_n$ and this is clearly connected.

Longer answer, from a more pedestrian viewpoint : a standard basis $\cal B$ for $\mathfrak g$ is $\lbrace h_i,a_i,b_i\rbrace_{1 \leq i \leq n} \cup \lbrace c_{i,j},d_{i,j},f_{i,j},g_{i,j}\rbrace_{1 \leq i where

(and $e_{ij}$ is the matrix all of whose coefficients are zero except for the $(i,j)$-th which is $1$).

Now let $\mathfrak i$ be an ideal of $\mathfrak g$. Then $\mathfrak i$ is stable by $ad(x)$ for any $x\in {\mathfrak g}$. In particular, if $H$ is the vector space spanned by
$h_1,h_2, \ldots ,h_n$, then $\mathfrak i$ is stable by all the $ad(h)$, for
$h \in H$. Now those $ad(h)$ are simultaneously diagonalized by the basis we have just described above. We deduce that $\mathfrak i=H' \oplus {\sf Vect}(F)$, where $H'$ is a linear subspace of $H$ and $F$ is a subset of ${\cal B}'={\cal B} \setminus \lbrace h_1,h_2, \ldots ,h_n \rbrace$. Let us now consider the root spaces

Each root space $R$ is a Lie subalgebra of $\mathfrak g$ isomorphic to ${\mathfrak sl}_2$. It easily follows that either ${\mathfrak i} \cap R=\lbrace 0 \rbrace$ or
$R \subseteq {\mathfrak i}$ for any $R$. Combining this with the decomposition of $\mathfrak i$ above, we see that $\mathfrak i$ is the direct sum of some root spaces, plus some subspace of $H$.

Suppose $F$ contains $a_i$ for some $i$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i,a_i,b_i)$. For $j>i$, $\mathfrak i$ also contains
$[a_i,g_{ij}]=c_{ij}$, $h_i-[c_{ij},d_{ij}]=h_j, [h_j,\frac{a_j}{2}]=a_j$. Similarly, for $k,
$\mathfrak i$ also contains $[a_i,-g_{ki}]=d_{ki}$, $h_i+[c_{ki},d_{ki}]=h_k, [h_k,\frac{a_k}{2}]=a_k$. So $\mathfrak i$ contains all the $a_i$, and we easily deduce that $\mathfrak i=A$.

Suppose $F$ contains $c_{ij}$ for some $i. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i-h_j,c_{ij},d_{ij})$. And $\mathfrak i$ also contains $h_i+h_j=[c_{ij},f_{ij}]$. So $\mathfrak i$ contains both $h_i$ and $h_j$, it also contains the root space
${\sf Vect}(h_i,a_i,b_i)$, so $\mathfrak i=A$ by the preceding paragraph.

Suppose $F$ contains $f_{ij}$ for some $i. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i+h_j,f_{ij},g_{ij})$. And $\mathfrak i$ also contains $c_{ij}=[f_{ij},b_{j}]$, so $\mathfrak i=A$ by the preceding paragraph.

Finally we see that $\mathfrak i=A$ unless $F$ is empty. In that case, $\mathfrak i$ is a linear subspace of $H$. For any $i$, $ad (a_i)$ must be zero on $\mathfrak i$, otherwise
$\mathfrak i$ would contain $a_i$. It is easily deduced that ${\mathfrak i}=\lbrace 0 \rbrace$,
so $\mathfrak g$ is indeed simple.